Question

Question: (a) Express the volume of the wedge in the first octant that is cut from the cylinder \({{\rm{y}}^{\rm{2}}}{\rm{ + }}{{\rm{z}}^{\rm{2}}}{\rm{ = 1}}\) by the planes \({\rm{y = x}}\) and \({\rm{x = 1}}\) as a triple integral. (b) Use either the Table of Integrals (on Reference Pages \({\rm{6--10}}\)) or a computer algebra system to find the exact value of the triple integral in part (a).

Short answer

1. The integral is\({\rm{V = }}\int_{\rm{0}}^{\rm{1}} {\int_{\rm{0}}^{\rm{x}} {\int_{\rm{0}}^{\sqrt {{\rm{1 - }}{{\rm{y}}^{\rm{2}}}} } {{\rm{dzdydx}}} } } \).
2. The volume is \(\frac{{\rm{1}}}{{\rm{4}}}{\rm{\pi - }}\frac{{\rm{1}}}{{\rm{3}}}\).

Step-by-step solution

Define integral

In mathematics, an integral assigns numbers to functions to express displacement, area, volume, and other concepts arising from linking infinitesimal data.

Explanation

1. Begin by identifying the integration region,

   \({\rm{E = \{ (x,y,z)|0}} \le {\rm{x}} \le {\rm{1,0}} \le {\rm{y}} \le {\rm{x,0}} \le {\rm{z}} \le \sqrt {{\rm{1 - }}{{\rm{y}}^{\rm{2}}}} {\rm{\} }}\)

   Then, the integral is,

   \({\rm{V = }}\int_{\rm{0}}^{\rm{1}} {\int_{\rm{0}}^{\rm{x}} {\int_{\rm{0}}^{\sqrt {{\rm{1 - }}{{\rm{y}}^{\rm{2}}}} } {{\rm{dzdydx}}} } } \)

Explanation

b. \({\rm{V = }}\int_{\rm{0}}^{\rm{1}} {\int_{\rm{0}}^{\rm{x}} {\int_{\rm{0}}^{\sqrt {{\rm{1 - }}{{\rm{y}}^{\rm{2}}}} } {{\rm{dzdydx}}} } } \)

With respect to\({\rm{z}}\), integrate.

\(\begin{array}{c}{\rm{V = }}\int_{\rm{0}}^{\rm{1}} {\int_{\rm{0}}^{\rm{x}} {{\rm{z|}}_{\rm{0}}^{\sqrt {{\rm{1 - }}{{\rm{y}}^{\rm{2}}}} }} } {\rm{dydx}}\\{\rm{V = }}\int_{\rm{0}}^{\rm{1}} {\int_{\rm{0}}^{\rm{x}} {\sqrt {{\rm{1 - }}{{\rm{y}}^{\rm{2}}}} } } {\rm{dydx}}\end{array}\)

Apply formula 30 from the Table of Integrals at the back of the book to integrate with respect to y, using a = 1.

\(\begin{array}{c}{\rm{V = }}\int_{\rm{0}}^{\rm{1}} {\frac{{\rm{1}}}{{\rm{2}}}\left( {{\rm{y}}\sqrt {{\rm{1 - }}{{\rm{y}}^{\rm{2}}}} {\rm{ + si}}{{\rm{n}}^{{\rm{ - 1}}}}{\rm{y}}} \right)} {\rm{|}}_{\rm{0}}^{\rm{x}}{\rm{dx}}\\{\rm{V = }}\int_{\rm{0}}^{\rm{1}} {\frac{{\rm{1}}}{{\rm{2}}}\left( {{\rm{x}}\sqrt {{\rm{1 - }}{{\rm{x}}^{\rm{2}}}} {\rm{ + si}}{{\rm{n}}^{{\rm{ - 1}}}}{\rm{x}}} \right)} {\rm{dx}}\end{array}\)

Dividing the integral into two parts and integrating each individually,

\({\rm{V = }}\int_{\rm{0}}^{\rm{1}} {\frac{{\rm{1}}}{{\rm{2}}}{\rm{x}}\sqrt {{\rm{1 - }}{{\rm{x}}^{\rm{2}}}} {\rm{dx + }}\int_{\rm{0}}^{\rm{1}} {\frac{{\rm{1}}}{{\rm{2}}}{\rm{si}}{{\rm{n}}^{{\rm{ - 1}}}}{\rm{x}}} } {\rm{dx}}\)

Using a u-substitution, integratethe left-hand part,

\(\begin{array}{c}\int_{\rm{0}}^{\rm{1}} {\frac{{\rm{1}}}{{\rm{2}}}{\rm{x}}\sqrt {{\rm{1 - }}{{\rm{x}}^{\rm{2}}}} {\rm{dx}}} \\{\rm{u = 1 - }}{{\rm{x}}^{\rm{2}}}\\{\rm{du = - 2xdx}}\\\int_{\rm{1}}^{\rm{0}} {\frac{{{\rm{ - 1}}}}{{\rm{4}}}\sqrt {\rm{u}} {\rm{du = }}\int_{\rm{0}}^{\rm{1}} {\frac{{\rm{1}}}{{\rm{4}}}\sqrt {\rm{u}} } {\rm{du}}} \\\frac{{\rm{1}}}{{\rm{6}}}{{\rm{u}}^{{\rm{3/2}}}}{\rm{|}}_{\rm{0}}^{\rm{1}}{\rm{ = }}\frac{{\rm{1}}}{{\rm{6}}}\end{array}\)

Apply formula 87 from the Table of Integrals to the right-hand portion to integrate it,

\(\begin{array}{l}\int_{\rm{0}}^{\rm{1}} {\frac{{\rm{1}}}{{\rm{2}}}{\rm{si}}{{\rm{n}}^{{\rm{ - 1}}}}{\rm{xdx}}} \\\frac{{\rm{1}}}{{\rm{2}}}{\rm{(x \times si}}{{\rm{n}}^{{\rm{ - 1}}}}{\rm{x + }}\sqrt {{\rm{1 - }}{{\rm{x}}^{\rm{2}}}} {\rm{)|}}_{\rm{0}}^{\rm{1}}{\rm{ = }}\frac{{\rm{1}}}{{\rm{4}}}{\rm{\pi - }}\frac{{\rm{1}}}{{\rm{2}}}\end{array}\)

To get the entire volume, add the two parts together,

\(\begin{array}{c}{\rm{V = }}\frac{{\rm{1}}}{{\rm{6}}}{\rm{ + }}\left( {\frac{{\rm{1}}}{{\rm{4}}}{\rm{\pi - }}\frac{{\rm{1}}}{{\rm{2}}}} \right)\\{\rm{ = }}\frac{{\rm{1}}}{{\rm{4}}}{\rm{\pi - }}\frac{{\rm{1}}}{{\rm{3}}}\end{array}\)

Therefore, the volume is \(\frac{{\rm{1}}}{{\rm{4}}}{\rm{\pi - }}\frac{{\rm{1}}}{{\rm{3}}}\).